Question

$\left(B\right)\stackrel{NaB{H}_4}{\leftarrow }C{H}_3{N}_3\stackrel{LAH}{\to }\left(A\right)$

The products (A) and (B) are:

• A. $C{H}_{3}N{H}_2$, $C{H}_{3}N{H}_2$
• B. $C{H}_{3}N{H}_2$, No reaction
• C. No reaction, $C{H}_{3}N{H}_2$
• D. No reaction, No reaction

Answer 1

The correct option is A $C{H}_{3}N{H}_2$, $C{H}_{3}N{H}_2$

$NaB{H}_4$ and $LiAl{H}_4$ both act as sources of nucleophilic hydride and therefore are reactive towards electrophilic species. In the reaction with $C{H}_3{N}_3$ they both form $C{H}_{3}N{H}_2$.

$C{H}_3-N=N^{+}=N^{-}+N{a}^{+}B{H}_4^{-}\to C{H}_3-N^{-}-B{H}_4^{+}+N\equiv N$

$C{H}_3-N^{-}-B{H}_4^{+}\to C{H}_{3}N{H}_2$